Problem: Complete the square to solve for $x$. $x^{2}-10x+21 = 0$
Begin by moving the constant term to the right side of the equation. $x^2 - 10x = -21$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-10$ , half of it would be $-5$ , and squaring it gives us ${25}$ $x^2 - 10x { + 25} = -21 { + 25}$ We can now rewrite the left side of the equation as a squared term. $( x - 5 )^2 = 4$ Take the square root of both sides. $x - 5 = \pm2$ Isolate $x$ to find the solution(s). $x = 5\pm2$ So the solutions are: $x = 7 \text{ or } x = 3$ We already found the completed square: $( x - 5 )^2 = 4$